Biết \(tan\)=\(\dfrac{1}{3}\). Gía trị công thức A= \(\dfrac{4sin\alpha+5cos\alpha}{2sin\alpha-3cos\alpha}\) bằng bao nhiêu?
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28 tháng 3 2022
\(A=\dfrac{\dfrac{4sin\alpha}{sin\alpha}+\dfrac{5cos\alpha}{sin\alpha}}{\dfrac{2sin\alpha}{sin\alpha}-\dfrac{3cos\alpha}{sin\alpha}}\)
\(A=\dfrac{4+5cot\alpha}{2-3cot\alpha}\)
Biết cotα=\(\dfrac{1}{2}\) nên ta có:
\(A=\dfrac{4+5\cdot\dfrac{1}{2}}{2-3\cdot\dfrac{1}{2}}\)
\(A=\dfrac{4+\dfrac{5}{2}}{2-\dfrac{3}{2}}\)
A= 13
26 tháng 3 2022
\(\dfrac{3sin\alpha-4cos\alpha}{2sin\alpha+3cos\alpha}=\dfrac{\dfrac{3sin\alpha}{cos\alpha}-\dfrac{4cos\alpha}{cos\alpha}}{\dfrac{2sin\alpha}{cos\alpha}+\dfrac{3cos\alpha}{cos\alpha}}=\dfrac{3tan\alpha-4}{2tan\alpha+3}\)
Biết tanα=\(-\dfrac{1}{4}\) nên ta có:
\(\dfrac{3\cdot\dfrac{-1}{4}-4}{2\cdot\dfrac{-1}{4}+3}=\dfrac{-\dfrac{3}{4}-4}{-\dfrac{1}{2}+3}=\dfrac{-19}{10}\)
PN
0
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 5 2019
\(\frac{2sina+3cosa}{4sina-5cosa}=\frac{\frac{2sina}{cosa}+\frac{3cosa}{cosa}}{\frac{4sina}{cosa}-\frac{5cosa}{cosa}}=\frac{2tana+3}{4tana-5}=\frac{6+3}{12-5}=\frac{9}{7}\)
\(\frac{3sina-2cosa}{5sina+4cos^3a}=\frac{\frac{3sina}{cosa}-\frac{2cosa}{cosa}}{\frac{5sina}{cosa}+\frac{4cos^3a}{cosa}}=\frac{3tana-2}{5tana+4cos^2a}=\frac{3tana-2}{5tana+\frac{4}{1+tan^2a}}=\frac{9-2}{15+\frac{4}{10}}=\frac{5}{11}\)
10 tháng 12 2020
Ta có: \(tan\alpha=2\Leftrightarrow\dfrac{sin\alpha}{cos\alpha}=2\Leftrightarrow sin\alpha=2cos\alpha\)
A = \(\dfrac{16cos^2\alpha+6cos^2\alpha}{20cos^2\alpha-2cos^2\alpha}=\dfrac{22cos^2\alpha}{18cos^2\alpha}=\dfrac{11}{9}\)
28 tháng 10 2023
Cách 1:
Ta có: \(tan\alpha=\sqrt{2}\Rightarrow\left\{{}\begin{matrix}\dfrac{sin\alpha}{cos\alpha}=\sqrt{2}\\1+\left(\sqrt{2}\right)^2=\dfrac{1}{cos^2\alpha}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=\sqrt{2}\cdot cos\alpha\\cos^2\alpha=\dfrac{1}{3}\end{matrix}\right.\)
\(P=\dfrac{sin\alpha-cos\alpha}{sin^3\alpha+3cos^3\alpha+2sin\alpha}\)
\(=\dfrac{\sqrt{2}\cdot cos\alpha-cos\alpha}{\left(\sqrt{2}\cdot cos\alpha\right)^3+3cos^3\alpha+2\cdot\sqrt{2}\cdot cos\alpha}\)
\(=\dfrac{cos\alpha\left(\sqrt{2}-1\right)}{2\sqrt{2}\cdot cos^3\alpha+3cos^3\alpha+2\sqrt{2}\cdot cos\alpha}\)
\(=\dfrac{cos\alpha\left(\sqrt{2}-1\right)}{cos\alpha\left(2\sqrt{2}\cdot cos^2\alpha+3cos^2\alpha+2\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{2}-1}{2\sqrt{2}\cdot cos^2\alpha+3cos^2\alpha+2\sqrt{2}}\)
Thay \(cos^2\alpha=\dfrac{1}{3}\) vào \(P\) ta có:
\(P=\dfrac{\sqrt{2}-1}{2\sqrt{2}\cdot\dfrac{1}{3}+3\cdot\dfrac{1}{3}+2\sqrt{2}}=\dfrac{\sqrt{2}-1}{1+\dfrac{8}{3}\sqrt{2}}\)
\(=\dfrac{3\left(\sqrt{2}-1\right)}{3\left(1+\dfrac{8}{3}\sqrt{2}\right)}=\dfrac{3\left(\sqrt{2}-1\right)}{3+8\sqrt{2}}\)
\(=\dfrac{3\left(\sqrt{2}-1\right)}{3+2^3\sqrt{2}}=\dfrac{a\left(\sqrt{b}-1\right)}{a+b^3\sqrt{b}}\)
\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\Rightarrow a+b=5\)
Chọn đáp án A.
28 tháng 10 2023
Cách 2:
\(P=\dfrac{sin\alpha-cos\alpha}{sin^3\alpha+3cos^3\alpha+2sin\alpha}=\dfrac{\left(sin\alpha-cos\alpha\right)\div cos^3\alpha}{\left(sin^3\alpha+3cos^3\alpha+2sin\alpha\right)\div cos^3\alpha}\)
\(=\dfrac{\dfrac{sin\alpha}{cos^3\alpha}-\dfrac{1}{cos^2\alpha}}{\dfrac{sin^3\alpha}{cos^3\alpha}+3+2\cdot\dfrac{sin\alpha}{cos^3\alpha}}=\dfrac{\dfrac{sin\alpha}{cos\alpha}\cdot\dfrac{1}{cos^2\alpha}-\dfrac{1}{cos^2\alpha}}{tan^3\alpha+3+2\cdot\dfrac{sin\alpha}{cos\alpha}\cdot\dfrac{1}{cos^2\alpha}}\)
\(=\dfrac{tan\alpha\cdot\left(1+tan^2\alpha\right)-\left(1+tan^2\alpha\right)}{tan^3\alpha+3+2tan\alpha\cdot\left(1+tan^2\alpha\right)}\)
Thay \(tan\alpha=\sqrt{2}\) vào ta có:
\(P=\dfrac{\sqrt{2}\cdot\left[1+\left(\sqrt{2}\right)^2\right]-\left[1+\left(\sqrt{2}\right)^2\right]}{\left(\sqrt{2}\right)^3+3+2\sqrt{2}\cdot\left[1+\left(\sqrt{2}\right)^2\right]}=\dfrac{3\sqrt{2}-3}{2\sqrt{2}+3+6\sqrt{2}}\)
\(=\dfrac{3\left(\sqrt{2}-1\right)}{3+8\sqrt{2}}=\dfrac{3\left(\sqrt{2}-1\right)}{3+2^3\sqrt{2}}=\dfrac{a\left(\sqrt{b}-1\right)}{a+b^3\sqrt{b}}\)
\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\Rightarrow a+b=3+2=5\)
Chọn đáp án A
28 tháng 3 2022
\(cot\alpha=\dfrac{1}{2}tacó:\)
\(cot\alpha=\dfrac{cos\alpha}{sin\alpha}\)
\(A=\dfrac{4sin\alpha+5cos\alpha}{2sin\alpha-3cos\alpha}\)
\(A=\dfrac{\dfrac{4sin\alpha}{sin\alpha}+\dfrac{5cos\alpha}{sin\alpha}}{\dfrac{2sin\alpha}{sin\alpha}-\dfrac{3cos\alpha}{sin\alpha}}\)
\(A=\dfrac{4+5cot\alpha}{2-3cot\alpha}\)
\(A=\dfrac{4+5\left(\dfrac{1}{2}\right)}{2-3\left(\dfrac{1}{2}\right)}\)
\(A=13\)
BU
7 tháng 4 2022
\(A=\dfrac{4\sin\alpha+5\cos\alpha}{2\sin\alpha-3\cos\alpha}\)
\(A=\dfrac{\dfrac{4\sin\alpha}{\sin\alpha}+\dfrac{5\cos\alpha}{\sin\alpha}}{\dfrac{2\sin\alpha}{\sin\alpha}-\dfrac{3\cos\alpha}{\sin\alpha}}\)
\(A=\dfrac{4+5\cot\alpha}{2-3\cot\alpha}\)
Thay cot α= \(\dfrac{1}{2}\) vào A, ta có:
\(A=\dfrac{4+5\times\dfrac{1}{2}}{2-3\times\dfrac{1}{2}}\)
\(A=\dfrac{4+\dfrac{5}{2}}{2-\dfrac{3}{2}}\)
\(A=\dfrac{13}{\dfrac{2}{\dfrac{1}{2}}}\)
A=13
\(\dfrac{4sin\alpha+5cos\alpha}{2sin\alpha-3cos\alpha}=\dfrac{\dfrac{4sin\alpha}{cos\alpha}+\dfrac{5cos\alpha}{cos\alpha}}{\dfrac{2sin\alpha}{cos\alpha}-\dfrac{3cos\alpha}{cos\alpha}}=\dfrac{4tan\alpha+5}{2tan\alpha-3}\)
Biết \(tan\)=\(\dfrac{1}{3}\) nên ta có:
\(\dfrac{4\times\dfrac{1}{2}+5}{2\times\dfrac{1}{2}-3}=\dfrac{2+5}{2-3}=\dfrac{7}{-2}=\dfrac{-7}{2}\)